What is the solution for 25y^{''}+10y=0 ?

The solution for 25y^{''}+10y=0 is y=c_{1}cos(sqrt(2/5)t)+c_{2}sin(sqrt(2/5)t)

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25y^{''}+10y=0

25 y^{^{''}} + 10 y = 0

y = c_{1} cos (\frac{2}{5} t) + c_{2} sin (\frac{2}{5} t)

Solution steps

25 y^{''} (t) + 10 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{2}{5} t) + c_{2} sin (\frac{2}{5} t)

y = c_{1} cos (\frac{2}{5} t) + c_{2} sin (\frac{2}{5} t)

\frac{d}{d x} (- ln (4 x))

\int (- 8 sec (x) tan (x) - 3 sec^{2} (x)) d x

y^{^{''}} + 5 y^{^{'}} - 6 y = (3 x - 1) e^{x}

e^{x^{4}}

x + y = 7

What is the solution for 25y^{''}+10y=0 ?
The solution for 25y^{''}+10y=0 is y=c_{1}cos(sqrt(2/5)t)+c_{2}sin(sqrt(2/5)t)